Divergent succession, but with convergent sum average.

Question

An example of a sequence $a_n$ such that: $$a_n\rightarrow\pm\infty$$ but $$b_n=\frac{\sum_{k=1}^{n}a_k}{n}$$ converge.

Answer 1

Assume without loss of generality that $a_n > 0$ always. Suppose that $b_n \to L$ a finite limit. Since $a_n \to \infty$, there exists some $N$ such that for all $n \geq N$, we have that $a_n > 2L$. But then for $n > 2N$, $$ \sum_{k = 1}^{n} a_n > 0 + \sum_{k = N}^{n} 2L > Ln.$$ So in particular, the average of these terms will always be more than $L$, contrary to the claim. This is a contradiction. $\diamondsuit$

More generally, it's true that if $a_n \to L$, then the average value of $a_n$ also converges to $L$.

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To match the edit, consider $$ a_n = (-1)^n \log n. $$